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x+2squarerootx-3=0. solve for x?
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\[x+2\sqrt{x}-3=0\] \[2\sqrt{x}=3-x\] square to get \[4x=(3-x)^2\] \[4x=9-6x+x^2\] \[x^2-10x+9=0\] solve for x get \[(x-1)(x-10)=0\] \[x=1\] or \[x=10\] butyou are not done here because you have to check your answers.
ok that was way wrong \[x^2-10x+9=0\] \[(x-1)(x-9)=0\] \[x=1\] or \[x=9\] now it is right
now check because there is no guarantee they work \[1+2\sqrt{1}-3=0\] \[1+2-3=0\] that one works \[9+2\sqrt{9}-3=0\] \[9+3-3=0\] that one does not work. only solution is \[x=1\]
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